laws-of-motion.pdf

genius PHYSICS

by Pradeep Kshetrapal

Newton’s Laws of Motion 1

1

R sin



R
R cos

4.1 Point Mass.
(1) An object can be considered as a point object if during motion in a given time, it covers distance much
greater than its own size.
(2) Object with zero dimension considered as a point mass.
(3) Point mass is a mathematical concept to simplify the problems.

4.2 Inertia.
(1) Inherent property of all the bodies by virtue of which they cannot change their state of rest or uniform
motion along a straight line by their own is called inertia.
(2) Inertia is not a physical quantity, it is only a property of the body which depends on mass of the body.
(3) Inertia has no units and no dimensions
(4) Two bodies of equal mass, one in motion and another is at rest, possess same inertia because it is a
factor of mass only and does not depend upon the velocity.

4.3 Linear Momentum.
(1) Linear momentum of a body is the quantity of motion contained in the body.
(2) It is measured in terms of the force required to stop the body in unit time.
(3) It is measured as the product of the mass of the body and its velocity i.e., Momentum = mass ×
velocity.
If a body of mass m is moving with velocity v then its linear momentum p is given by p  m v
(4) It is a vector quantity and it’s direction is the same as the direction of velocity of the body.
(5) Units : kg-m/sec [S.I.], g-cm/sec [C.G.S.]
(6) Dimension : [MLT

1

]

(7) If two objects of different masses have same momentum, the lighter body
possesses greater velocity.

v
p = constant

p  m 1v1  m 2 v 2 = constant



v1 m 2

v 2 m1

i.e. v 

(8) For a given body p  v
p

m=
constant

v

1
m

[As p is constant]

m

(9) For different bodies at same velocities p  m
p

v = constant

m

2

2 Newton’s Laws of Motion

2
4.4 Newton’s First Law.

A body continue to be in its state of rest or of uniform motion along a straight line, unless it is acted upon
by some external force to change the state.
(1) If no net force acts on a body, then the velocity of the body cannot change i.e. the body cannot
accelerate.
(2) Newton’s first law defines inertia and is rightly called the law of inertia. Inertia are of three types :
Inertia of rest, Inertia of motion, Inertia of direction
(3) Inertia of rest : It is the inability of a body to change by itself, its state of rest. This means a body at
rest remains at rest and cannot start moving by its own.
Example : (i) A person who is standing freely in bus, thrown backward, when bus starts suddenly.
When a bus suddenly starts, the force responsible for bringing bus in motion is also transmitted to lower
part of body, so this part of the body comes in motion along with the bus. While the upper half of body (say
above the waist) receives no force to overcome inertia of rest and so it stays in its original position. Thus there
is a relative displacement between the two parts of the body and it appears as if the upper part of the body has
been thrown backward.

Note : 

If the motion of the bus is slow, the inertia of motion will be transmitted to the body of
the person uniformly and so the entire body of the person will come in motion with the bus and
the person will not experience any jerk.

(ii) When a horse starts suddenly, the rider tends to fall backward on account of inertia of rest of upper
part of the body as explained above.
(iii) A bullet fired on a window pane makes a clean hole through it while a stone breaks the whole window
because the bullet has a speed much greater than the stone. So
its time of contact with glass is small. So in case of bullet the
motion is transmitted only to a small portion of the glass in that
small time. Hence a clear hole is created in the glass window,
while in case of ball, the time and the area of contact is large.
During this time the motion is transmitted to the entire window,
thus creating the cracks in the entire window.
Cracks by the ball
Hole by the bullet
(iv) In the arrangement shown in the figure :
(a) If the string B is pulled with a sudden jerk then it will
experience tension while due to inertia of rest of mass M this force will not be transmitted
to the string A and so the string B will break.
A
(b) If the string B is pulled steadily the force applied to it will be transmitted from
string B to A through the mass M and as tension in A will be greater than in B by Mg
M
(weight of mass M) the string A will break.
(v) If we place a coin on smooth piece of card board covering a glass and strike the
B
card board piece suddenly with a finger. The cardboard slips away and the coin falls into the
glass due to inertia of rest.
(vi) The dust particles in a durree falls off when it is beaten with a stick. This is
because the beating sets the durree in motion whereas the dust particles tend to remain at rest and hence
separate.
(4) Inertia of motion : It is the inability of a body to change itself its state of uniform motion i.e., a body
in uniform motion can neither accelerate nor retard by its own.
Example : (i) When a bus or train stops suddenly, a passenger sitting inside tends to fall forward. This is
because the lower part of his body comes to rest with the bus or train but the upper part tends to continue its
motion due to inertia of motion.

genius PHYSICS

by Pradeep Kshetrapal

Newton’s Laws of Motion 3

3

(ii) A person jumping out of a moving train may fall forward.
(iii) An athlete runs a certain distance before taking a long jump. This is because velocity acquired by
running is added to velocity of the athlete at the time of jump. Hence he can jump over a longer distance.
(5) Inertia of direction : It is the inability of a body to change by itself direction of motion.
Example : (i) When a stone tied to one end of a string is whirled and the string breaks suddenly, the stone
flies off along the tangent to the circle. This is because the pull in the string was forcing the stone to move in a
circle. As soon as the string breaks, the pull vanishes. The stone in a bid to move along the straight line flies off
tangentially.
(ii) The rotating wheel of any vehicle throw out mud, if any, tangentially, due to directional inertia.
(iii) When a car goes round a curve suddenly, the person sitting inside is thrown outwards.
Sample problem based on Newton’s first law

Problem 1.

When a bus suddenly takes a turn, the passengers are thrown outwards because of
[AFMC 1999; CPMT 2000, 2001]

(a) Inertia of motion

(b) Acceleration of motion

(c) Speed of motion

(d) Both (b) and (c)

Solution : (a)

Problem 2.

A person sitting in an open car moving at constant velocity throws a ball vertically up into air. The ball fall
[EAMCET (Med.) 1995]

Solution : (d)

(a) Outside the car

(b) In the car ahead of the person

(c) In the car to the side of the person

(d) Exactly in the hand which threw it up

Because the horizontal component of velocity are same for both car and ball so they cover equal horizontal
distances in given time interval.

4.5 Newton’s Second Law.
(1) The rate of change of linear momentum of a body is directly proportional to the external force applied
on the body and this change takes place always in the direction of the applied force.



(2) If a body of mass m, moves with velocity v then its linear momentum can be given by p  m v and if


y and z axis.
(iii) Velocity dependent force : Viscous force (6 rv)
Force on charged particle in a magnetic field (qvB sin  )
(13) Central force : If a position dependent force is always directed towards or away from a fixed point it is
said to be central otherwise non-central.

 d

dv  dm
(5) If m is not constant F  (m v )  m
v
dt
dt
dt
(6) If force and acceleration have three component along x. nuclear force is strongest while gravitational
force weakest.
.
(12) Variable or dependent force :
(i) Time dependent force : In case of impulse or motion of a charged particle in an alternating electric field
force is time dependent.
(ii) Position dependent force : Gravitational force between two bodies
Force between two charged particles 
or
Gm 1 m 2
r2
q1 q 2
4 0 r 2
.
Example : Positive force – Force between two similar charges
Negative force – Force between two opposite charges
(9) Out of so many natural forces.
Example : Motion of earth around the sun. Fy  ma y . for distance 10 15 metre. Fz  ma z
(7) No force is required to move a body uniformly along a straight line. the force is said to be conservative otherwise non conservative.
F  0
F  ma
(As a  0 )
(8) When force is written without direction then positive force means repulsive while negative force
means attractive. Scattering of -particles from a
nucleus.
F
Sun
F
+
Earth
Nucleus
–
Electron
F
+
Nucleus
+
particle
(14) Conservative or non conservative force : If under the action of a force the work done in a round trip is
zero or the work is path independent. Fnuclear  Felectromag netic  Fgravitational
(10) Ratio of electric force and gravitational force between two electron Fe / Fg  10 43  Fe  Fg
(11) Constant force : If the direction and magnitude of a force is constant. Motion of electron in an atom. then


F  Fx ˆi  Fy ˆj  Fz kˆ and a  a x ˆi  a y ˆj  a z kˆ
From above it is clear that Fx  ma x .genius PHYSICS
by Pradeep Kshetrapal
Newton’s Laws of Motion 5
5


(4) F  m a formula is valid only if force is changing the state of rest or motion and the mass of the body is
constant and finite. It is said to be a constant force.

then they are said to be concurrent. rope or chain against pulling (applied) force is
called the tension.
F2
B
C
F3
F1
A
. This resistive force increases with
change in length.7 Equilibrium of Concurrent Force.
(4) Mathematically for equilibrium
F
net
 0 or
F
x
0.
F = – Kx
x
4.
T=F
(iv) Spring force : Every spring resists any attempt to change its length.
(ii) Reaction or Normal force : When a body is placed on a rigid surface.
F
y
0.
R
R

 mg
mg
mg cos
(iii) Tension : The force exerted by the end of taut string. elastic force. the body experiences a force
which is perpendicular to the surfaces in contact.
(15) Common forces in mechanics :
(i) Weight : Weight of an object is the force with which earth attracts it. It is also called the force of gravity
or the gravitational force. The direction of tension is so as to pull the body.
(2) A body..6
6 Newton’s Laws of Motion
6
Example : Conservative force : Gravitational force.
(1) If all the forces working on a body are acting on the same point. when there is no change in
the state of rest or of uniform motion along a straight line. electric force. Spring force is given by F  Kx . Then force is called ‘Normal force’ or ‘Reaction’. if they can be represented completely by three sides of
a triangle taken in order. viscous force. under the action of concurrent forces.
F
z
0
(5) Three concurrent forces will be in equilibrium. where x is the change in length and K is the spring
constant (unit N/m). is said to be in equilibrium.
(3) The necessary condition for the equilibrium of a body under the action of concurrent forces is that the
vector sum of all the forces acting on the body must be zero.
Non conservative force : Frictional force.

A spring balance A shows a reading of 2 kg.
A weightless ladder.
Choose the correct magnitude from the following
[CBSE PMT 1998]
(a) 175 lb
Solution: (c)
(b) 100 lb
(c) 70 lb
Since the system is in equilibrium therefore
F
 0 and
x
(d) 150 lb
F
y
 0  F  R 2 and W  R1
Now by taking the moment of forces about point B. If the rope PQ makes angle  with the vertical then
the tension in the string PQ is
P
(a) F sin
(b) F / sin

(c) F cos
v
(d) F / cos 
Solution: (b)
M
From the figure
T
For horizontal equilibrium
T sin  F
 T 
F
sin 
F
Q
T cos

F
T sin
mg
Problem 11. and it is pulled horizontally with a force F.
B
Problem 12.(BC)  W.(20 sin 60)  W(4 cos 60)  R1 (20 cos 60)
As R1
10 3 F  2W  10 R1
8W
 F

8  150
10 3
 W
 70 lb
4 ft
Wall
16 ft
R1
F
B
D
W
60o
A
E
C
10 3
Problem 10. A mass M is suspended by a rope from a rigid support at P as shown in the figure.(EC)  R1 ( AC)
[from the figure EC= 4 cos 60]
R2
F. Another rope is tied at
the end Q. A horizontal force is needed to prevent it from slipping.
F. pulley P1 is movable and pulley P2 is fixed. In the following diagram. Then [IIT-JEE 1
(a) The reading of the balance A will be more than 2 kg
(b) The reading of the balance B will be less than 5 kg
A
(c) The reading of the balance A will be less than 2 kg. The value of angle  will be
(a) 60o

P1
P2
(b) 30o
(c) 45o
(d) 15o
Solution: (b)
W
W
Free body diagram of pulley P1 is shown in the figure
For horizontal equilibrium T1 cos  T2 cos  T1  T2
and T1  T2  W
T1 sin
T2 sin
T1

T2

T1 cos
T2 cos
W
.8
8 Newton’s Laws of Motion
8
Problem 9. and that of B will be 5 kg. Another
balance B shows a reading of 5 kg. 20 ft long rests against a frictionless wall at an angle of 60 o with the horizontal. A 150
pound man is 4 ft from the top of the ladder. when a beaker full of liquid is placed in its pan. The two balances are
arranged such that the Al – block is completely immersed inside the liquid as shown in the figure. and that of B will be
more than 5 kg
(d) The reading of balance A will be 2 kg.
Solution: (c)
m
Due to buoyant force on the aluminium block the reading of spring balance
A will be less than 2 kg but it increase the reading of balance B. when an aluminium block is suspended from it.

(2) Forces in nature always occurs in pairs.
R sin
(vi) It is difficult to walk on sand or ice. the second body also exerts an equal and opposite force
on the first.
(5) If F AB = force exerted on body A by body B (Action) and F BA = force exerted on body B by body A
(Reaction)
Then according to Newton’s third law of motion F AB   F BA
(6) Example : (i) A book lying on a table exerts a force on the table which is equal to the weight of the
book. A cricket ball of mass 150 gm is moving with a velocity of 12 m/s and is hit by a bat so that the ball is
turned back with a velocity of 20 m/s.01s on the ball.
mg
As the system is at rest.
(1) When a body exerts a force on any other body. 01
4. The force of blow acts for 0.
(vii) Driving a nail into a wooden block without holding the block is difficult. the body will always remain in equilibrium. t  0. A single isolated force is not possible.
R
The table supports the book. by exerting an equal force on the book.01 sec
Force exerted by the bat on the ball F 
m [v 2  v1 ] 0 .genius PHYSICS
by Pradeep Kshetrapal
Newton’s Laws of Motion 11
11
2
F
2
2
2
   F  F  2 F cos 
3
 
F2
 2 F 2  2 F 2 cos 
9

17 2
  17 
1   17 
F  2 F 2 cos   cos  
 or   cos 

9
18


 18 
Problem 21. applying a force also experiences a force of equal magnitude but in opposite direction.
(3) Any agent. the bullet moves forward (action). The component of reaction in horizontal direction makes the person move
forward.
To every action.15kg .8 Newton’s Third Law.
(ii) Swimming is possible due to third law of motion. How can you get off if no horizontal force is exerted by pushing
against the surface
.
(v) While walking a person presses the ground in the backward direction (action)
by his feet. 15[20  (12)]

= 480 Newton
t
0 . This is the
force of reaction. Therefore force of action and
reaction must be equal and opposite.e. The gun recoils backward (reaction)
(iv) Rebounding of rubber ball takes place due to third law of motion.
(4) Action and reaction never act on the same body.
(iii) When a gun is fired. If it were so the total force on a body would have
always been zero i. The
force applied by the agent is called ‘Action’ and the counter force experienced by it is called ‘Reaction’. The ground pushes the person in forward direction with an equal force
(reaction). The average force
exerted by the bat on the ball is
(a) 480 N
Solution : (a)
v1  12m / s
(b) 600 N
(c) 500 N
(d) 400 N
and v 2  20 m / s [because direction is reversed]
m  150 gm  0. net force on it is zero.

R
R cos
Sample problem based on Newton’s third law
Problem 22. there is always an equal (in magnitude) and opposite (in direction) reaction. This is the force of action. You are on a frictionless horizontal plane.

earth can not be considered as an inertial frame but for this purpose
the sun may be assumed to be an inertial frame.10 Impulse.s 1 (C. velocity.
(d) Ideally no inertial frame exist in universe.
(i) Inertial frame of reference :
(a) A frame of reference which is at rest or which is moving with a uniform velocity along a straight line is
called an inertial frame of reference.
If we plot a graph between force and time.
(5) Dimension : [ MLT 1 ]
(6) Units : Newton-second or Kg-m.
(ii) Non inertial frame of reference :
(a) Accelerated frame of references are called non-inertial frame of reference.
(3) Frame of reference are of two types : (i) Inertial frame of reference (ii) Non-inertial frame of
reference. For practical purpose a frame of reference may be
considered as inertial if it’s acceleration is negligible with respect to the acceleration of the object to be
observed.12
12 Newton’s Laws of Motion
12
Solution : (b)
(a) By jumping
(b) By splitting or sneezing
(c) By rolling your body on the surface
(d) By running on the plane
By doing so we can get push in backward direction in accordance with Newton’s third law of motion.G. In such case we measure the total effect of force.)
(7) Force-time graph : Impulse is equal to the area under F-t curve. lift which is moving upward or downward with some
acceleration.
Example : The lift at rest.
(f) To observe the motion of planets. of an object in this coordinate system.
(2) The reference frame is associated with a co-ordinate system and a clock to measure the position and
time of events happening in space.
acceleration etc.
An impulsive force does not remain constant.
4.
. earth can be considered as an inertial frame.
1
(4) Impulse is a vector quantity and its direction is same as that of force.I.
(b) In inertial frame of reference Newton’s laws of motion holds good. We can describe all the physical quantities like position.S.
(b) Newton’s laws of motion are not applicable in non-inertial frame of reference. the area under the curve and time axis gives the value of
impulse.9 Frame of Reference.
(e) To measure the acceleration of a falling apple.
(1) A frame in which an observer is situated and makes his observations is known as his ‘Frame of
reference’. plane which is taking off.
(c) Inertial frame of reference are also called unaccelerated frame of reference or Newtonian or Galilean
frame of reference.
Example : Car moving in uniform circular motion. it is called impulsive force.s 1 (S. car moving with constant
velocity on a straight road.
(3) I 
t2
t
F dt .
(1) When a large force works on a body for very small time interval. lift moving (up or down) with constant velocity.
(2) Impulse of a force is a measure of total effect of force. but changes first from zero to maximum and then from
maximum to zero.
4.)
and
Dyne-second or gm-cm.

The impulse of a force is equal to the change in momentum.
(iii) In jumping on sand (or water) the time of contact is increased due to
yielding of sand or water so force is decreased and we are not injured. then
impulse is
(a) 0.genius PHYSICS
by Pradeep Kshetrapal
Newton’s Laws of Motion 13
13

1
 Base  Height
2

1
Ft
2
Force
I  Area between curve and time axis
F
t
Time
(8) If Fav is the average magnitude of the force then
I

t2
t1
F dt  Fav

t2
t1
F
dt  Fav t
Fav

Impulse
dp
(9) From Newton’s second law F 
dt

t2
p2
or
F dt 
d p  I  p 2  p 1  p
t
1
t1
t
t2
t
p
1
i. which acts on it for 0.1 = 0.15  10 3 N .
(10) Examples : Hitting. kicking. catching.s
Impulse  force  time  50  10 5  3  1.1 sec.5  10 3 N .
(v) China wares are wrapped in straw or paper before packing. lesser force acts on his hands and his hands are saved from
getting hurt.
Sample problem based on Impulse
Problem 23. However if we
jump on cemented floor the motion stops in a very short interval of time resulting in
a large force due to which we are seriously injured.5  10 3 N . A force of 50 dynes is acted on a body of mass 5 g which is at rest for an interval of 3 seconds.
(iv) An athlete is advised to come to stop slowly after finishing a fast race. I 

F dt  Fav .s
. diving. jumping.
(i) In hitting or kicking a ball we decrease the time of contact so that large force acts on the ball producing
greater acceleration.5  10 3 N .s
(c) 1.s
(d) 2.2 N-s
 m a  t  0.
(ii) In catching a ball a player by drawing his hands backwards increases the
time of contact and so. So that time of stop increases
and hence force experienced by him decreases.3 N-s
(d) 1. t  p  constant
So if time of contact t is increased.e. collision etc.5 N-s
Solution : (c)
[AFMC 1999]
(b) 0.15  20  0.3 N-s
Problem 24.
This statement is known as Impulse momentum theorem.98  10 3 N . average force is decreased (or diluted) and vice-versa.1 N-s
Impulsive force  force time
(c) 0. The
impulsive force is
(a) 0. A ball of mass 150g moving with an acceleration 20m / s 2 is hit by a force.s
Solution : (c)
(b) 0.
In all these cases an impulse acts.

(3) Conservation of linear momentum is equivalent to Newton’s third law of motion.  constant



This equation shows that in absence of external force for a closed system the linear momentum of
individual particles may change but their sum remains unchanged with time..
4..

m 1 v1  m 2 v 2  constant.
Let m G  mass of gun.
(1) According to this law for a system of particles F 
dp
dt


In the absence of external force F  0 then p  constant
i.


p 1  p 2  constant.
or
m 1 v 1  m 2 v 2  m 3 v 3  .e. m B  mass of bullet. So net area is zero.
(2) Law of conservation of linear momentum is independent of frame of reference though linear
momentum depends on frame of reference.
(iii) Recoiling of a gun : For bullet and gun system. v B  velocity of bullet
Initial momentum of system = 0


Final momentum of system  m G v G  m B v B
By the law of conservation linear momentum

vG

vB
.
(ii) A person left on a frictionless surface can get away from it by blowing air out of his mouth or by
throwing some object in a direction opposite to the direction in which he wants to move.. the boat is pushed slightly away from the shore.14
14 Newton’s Laws of Motion
14
Problem 25. The
momentum acquired by the particle in time interval from zero to 8 second will be
(a) – 2 N-s
Force (N)
(b) + 4 N-s
(c) 6 N-s
(d) Zero
Solution : (d)
+2
2
–2
4
6
8
Time (s)
Momentum acquired by the particle is numerically equal to the area enclosed between the F-t curve and
time Axis.
If no external force acts on a system (called isolated) of constant mass.. for every action there is equal and opposite reaction which is Newton’s third law of motion.
For a system of two particles in absence of external force by law of conservation of linear momentum..
(4) Practical applications of the law of conservation of linear momentum
(i) When a man jumps out of a boat on the shore.11 Law of Conservation of Linear Momentum.e. The force-time (F – t) curve of a particle executing linear motion is as shown in the figure. For the given diagram area in a upper half is positive and in lower half is negative (and equal to
the upper half).
v G  velocity of gun... Hence the momentum acquired by the particle will be zero. the force exerted by trigger will be internal so the
momentum of the system remains unaffected.
Differentiating above with respect to time




d v1
dv 2
m1
 m2
 0  m 1 a1  m 2 a 2  0
dt
dt

 F1  F 2  0
F 2  F1
i.  constant.
p  p 1  p 2  p 3  . the total momentum of the system
remains constant with time.

4.12 Free Body Diagram.
In this diagram the object of interest is isolated from its surroundings and the interactions between the
object and the surroundings are represented in terms of forces.
Example :
m1a

T

T

m1
m1g sin

T

a

a

m1

Free body
diagram of mass
m1

m2a

T
m2

m2





Free body
diagram of mass
m2

m2g sin

4.13 Apparent Weight of a Body in a Lift.
When a body of mass m is placed on a weighing machine which is placed in a lift,
then actual weight of the body is mg. This acts on a weighing machine which offers a
reaction R given by the reading of weighing machine. This reaction exerted by the
surface of contact on the body is the apparent weight of the body.

R

mg

Condition

Figure

Velocity

Acceleration

Reaction

Conclusion

LIFT
R

Lift is at rest

v=0

a=0

R – mg = 0
 R = mg

Apparent weight =
Actual weight

a=0

R – mg = 0
 R = mg

Apparent weight =
Actual weight

a<g

R – mg = ma
R = m(g + a)

Apparent weight >
Actual weight

a=g

R – mg = mg
R = 2mg

Apparent weight =
2 Actual weight

Spring Balance

mg
LIFT

Lift moving
upward or
downward with
constant velocity

R

v = constant

Spring Balance

mg
LIFT

Lift accelerating
upward at the
rate of 'a’

R

a

v = variable

Spring Balance

mg
LIFT

Lift accelerating
upward at the
rate of ‘g’

R

Spring Balance

mg

v = variable
g

18

18 Newton’s Laws of Motion

18

LIFT
R

Lift accelerating
downward at the
rate of ‘a’

a

v = variable

g

v = variable

a<g

mg – R = ma
 R = m(g – a)

Apparent weight <
Actual weight

a=g

mg – R = mg
R=0

Apparent weight =
Zero
(weightlessness)

mg – R = ma
R = mg – ma
R = – ve

Apparent weight
negative means the
body will rise from
the floor of the lift
and stick to the
ceiling of the lift.

Spring Balance

mg
LIFT
R

Lift accelerating
downward at the
rate of ‘g’

Spring Balance

mg
LIFT
R

Lift accelerating
downward at the
rate of a(>g)

v = variable

a>g

a>g

Spring Balance

mg

Sample problems based on lift
Problem 33. A man weighs 80kg. He stands on a weighing scale in a lift which is moving upwards with a uniform
acceleration of 5 m / s 2 . What would be the reading on the scale. (g  10 m / s 2 )
(a) 400 N
Solution : (c)

(b) 800 N

(c) 1200 N

(d) Zero

Reading of weighing scale  m(g  a)  80 (10  5)  1200 N

Problem 34. A body of mass 2 kg is hung on a spring balance mounted vertically in a lift. If the lift descends with an
acceleration equal to the acceleration due to gravity ‘g’, the reading on the spring balance will be
(a) 2 kg
Solution : (d)

(b) (4  g) kg

R  m(g  a)  (g  g)  0

(c) (2  g) kg

(d) Zero

[because the lift is moving downward with a = g]

Problem 35. In the above problem, if the lift moves up with a constant velocity of 2 m/sec, the reading on the balance
will
Be
Solution : (a)

(a) 2 kg

(b) 4 kg

R  mg  2 g Newton or 2 kg

(c) Zero

(d) 1 kg

[because the lift is moving with the zero acceleration]

Problem 36. If the lift in problem, moves up with an acceleration equal to the acceleration due to gravity, the reading
on the spring balance will be
(a) 2 kg
Solution : (d)

R  m(g  a)  m (g  g)

(b) (2  g) kg

(c) (4  g) kg

(d) 4 kg

[because the lift is moving upward with a =g]

 2mg R  2  2 g N  4 g N or 4 kg

Problem 37. A man is standing on a weighing machine placed in a lift, when stationary, his weight is recorded as 40 kg.
If the lift is accelerated upwards with an acceleration of 2 m / s  , then the weight recorded in the machine
will be (g  10 m / s 2 )
(a) 32 kg

[MP PMT 1994]

(b) 40 kg

(c) 42 kg

(d) 48 kg

genius PHYSICS

by Pradeep Kshetrapal

Newton’s Laws of Motion 19

19

Solution : (d)

R  m (g  a)  40(10  2)  480 N or 48 kg

Problem 38. An elevator weighing 6000 kg is pulled upward by a cable with an acceleration of 5ms 2 . Taking g to be
10ms 2 , then the tension in the cable is
(a) 6000 N
Solution : (d)

[Manipal MEE 1995]

(b) 9000 N

(c) 60000 N

(d) 90000 N

T  m (g  a)  6000 (10  5) T  90,000 N

Problem 39. The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform
acceleration ‘a’ is 3 : 2. The value of ‘a’ is (g- Acceleration due to gravity on the earth)
(a)
Solution : (b)

Problem 40. A 60 kg man stands on a spring scale in the lift. At some instant he finds, scale reading has changed from
60 kg to 50 kg for a while and then comes back to the original mark. What should we conclude
(a) The lift was in constant motion upwards
(b) The lift was in constant motion downwards
(c) The lift while in constant motion upwards, is stopped suddenly
(d) The lift while in constant motion downwards, is suddenly stopped
Solution : (c)

For retarding motion of a lift R  m(g  a) for downward motion

R  m(g  a) for upward motion
Since the weight of the body decrease for a while and then comes back to original value it means the lift
was moving upward and stops suddenly.

Problem 41. A bird is sitting in a large closed cage which is placed on a spring balance. It records a weight placed on a

spring balance. It records a weight of 25 N. The bird (mass = 0.5kg) flies upward in the cage with an
acceleration of 2m / s 2 . The spring balance will now record a weight of
[MP PMT 1999]
(a) 24 N

Solution : (b)

(b) 25 N

(c) 26 N

(d) 27 N

Since the cage is closed and we can treat bird cage and air as a closed (Isolated) system. In this condition
the force applied by the bird on the cage is an internal force due to this reading of spring balance will not
change.

Problem 42. A bird is sitting in a wire cage hanging from the spring balance. Let the reading of the spring balance be
W1 . If the bird flies about inside the cage, the reading of the spring balance is W2 . Which of the following
is true

Solution : (b)

(a) W1  W2

(b) W1  W2

(c) W1  W2

(d) Nothing definite can be predicted

In this problem the cage is wire-cage the momentum of the system will not be conserved and due to this
the weight of the system will be lesser when the bird is flying as compared to the weight of the same
system when bird is resting is W2  W1 .

For rotational equilibrium about point ‘O’
(X  W ) a  (Y  W1 ) b
…. It is
always errorless. Spring is stretched and the weight of the body can be measured by the reading of
spring balance R  W  mg
The mechanism of weighing machine is same as that of spring balance.
For rotational equilibrium about point ‘O’
(X  W2 ) a  (Y  W ) b
From (i).. (ii) and (iii)
True weight W  W1 W2
……(iii)
. a = b.13)]
(2) Physical balance : In physical balance actually we compare the mass of body in both the pans.
Y
Effect of frame of reference : If the physical balance is perfect
then there will be no effect of frame of reference (either inertial or non-inertial) on the measurement. X = Y
X
and the needle must in middle of the beam i.genius PHYSICS
by Pradeep Kshetrapal
Newton’s Laws of Motion 33
33
4.
(ii) False balance : When the masses of the pan are not equal
then balance shows the error in measurement.
a
b
A
B
O
(i) Perfect physical balance :
Weight of the pan should be equal i. Here
we does not calculate the absolute weight of the body.e. False balance may be
of two types
a
b
A
B
O
(a) If the beam of physical balance is horizontal (when the
pans are empty) but the arms are not equal
Y
X
X  Y and a  b
For rotational equilibrium about point ‘O’
Xa  Yb
……(i)
In this physical balance if a body of weight W is placed in pan X then to balance it we have to put a weight
W1 in pan Y.e.
Here X and Y are the mass of the empty pan.20 Spring Balance and Physical Balance.
(1) Spring balance : When its upper end is fixed with rigid support and body of mass m hung from its
lower end.(ii)
Now if the pans are changed then to balance the body we have to put a weight W2 in pan X.
R
Effect of frame of reference : In inertial frame of reference the reading of spring
balance shows the actual weight of the body but in non-inertial frame of reference reading of
spring balance increases or decreases in accordance with the direction of acceleration
m
[for detail refer Article (4.

e. What will be
the angle of inclination with vertical
(a) tan 1 (a / g)
Solution : (a)
(b) tan 1 (g / a)
[Orissa JEE 2003]
(c) cos 1 (a / g)
From the figure
(d) cos 1 (g / a)
a
a
tan  
g

  tan 1 a / g 
a

g
Problem 74. A plumb line is suspended from a ceiling of a car moving with horizontal acceleration of a.(ii)
X
From (i) and (ii)
True weight W 
b
a
W1  W2
2
Sample problems (Miscellaneous)
Problem 72. when placed in the other pan of a false balance. Then mass of
2
[Orissa JEE 2002]
. A perpendicular force of 5 N acts on it
for 4 sec. when placed in one pan and 18gm. A block of mass 5 kg is moving horizontally at a speed of 1.. 5  4  6m
S y uy t 
1
1
1 2
at  0  F / m  t 2  5 / 5 4 2  8m
2
2
2
 S  S x2  S y2  36  64  100  10 m
Problem 75..5 m/s.
S x  u x  t  1 .
Problem 73. A body weighs 8 gm. If
the beam is horizontal (when both the pans are empty).34
34 Newton’s Laws of Motion
34
(b) If the beam of physical balance is not horizontal (when the pans are empty) and the arms are equal
i.5 gm
(d) 15 gm
W1W2  8  18 =12 gm.
We have to put a weight W1 in Y Pan
For equilibrium X  W  Y  W1
….
For equilibrium X  W2  Y  W
B
O
Y
…. What will be the distance of the block from the point where the force started acting [Pb PMT 2002]
(a) 10 m
Solution : (a)
(b) 8 m
(c) 6 m
(d) 2 m
In the given problem force is working in a direction perpendicular to initial velocity.
X  Y and a  b
In this physical balance if a body of weight W is placed in X Pan then to balance it. The velocity of a body of rest mass m 0 is
this body is
3
c (where c is the velocity of light in vacuum). So the body will move
under the effect of constant velocity in horizontal direction and under the effect of force in vertical
direction.(i)
A
Now if pans are changed then to balance the body we have
to put a weight W2 in X Pan. the true weight of the body is
(a) 13 gm
Solution : (b)
(b) 12 gm
For given condition true weight =
(c) 15.

This resistance is represented by a single force and is called friction.
(3) Kinetic or dynamic friction : If the applied force is increased further and sets the body in motion.
(1) Static friction : The opposing force that comes into play when one body tends to move over the
surface of another.
(ii) If a body is at rest and no pulling force is acting on it.
Fk  R or Fk   k R where  k is called the coefficient of kinetic friction
.
(i) The magnitude of limiting friction between any two bodies in contact is directly proportional to the
normal reaction between them.
(2) Limiting friction : If the applied force is increased the force of static friction also increases. If the
applied force exceeds a certain (maximum) value.2 Types of Friction.
P
F
mg
(iii) Static friction is a self-adjusting force because it changes itself in accordance with the applied force.1 Introduction.
(i) Kinetic friction depends upon the normal reaction. force of
friction on it is zero. This maximum value of static
friction upto which body does not move is called limiting friction.
Fl  R or Fl   s R
(ii) Direction of the force of limiting friction is always opposite to the direction in which one body is at the
verge of moving over the other
(iii) Coefficient of static friction : (a)  s is called coefficient of static friction and defined as the ratio of
force of limiting friction and normal reaction  s 
F
R
(b) Dimension : [M 0 L0 T 0 ]
(c) Unit : It has no unit.
rough or smooth polished or non-polished.
the friction opposing the motion is called kinetic friction.
(d) Value of  s lies in between 0 and 1
(e) Value of  depends on material and nature of surfaces in contact that means whether dry or wet .
R
(i) If applied force is P and the body remains at rest then static
friction F = P.
(f) Value of  does not depend upon apparent area of contact.
If we slide or try to slide a body over a surface the motion is resisted by a bonding between the body and
the surface.
The force of friction is parallel to the surface and opposite to the direction of intended motion. the body starts moving. but the actual motion has yet not started is called
static friction.36
36 Newton’s Laws of Motion
36
5.
5.

g.
(b) Rolling friction : When objects such as a wheel (disc or ring).
It is a general misconception that friction always opposes the motion. For example :
. coefficient of kinetic friction is always less than coefficient of static friction. It would have the dimensions of length and would be measured in
metre. Thus we require more
force to start a motion than to maintain it against friction.4 Friction is a Cause of Motion. No doubt friction opposes the
motion of a moving body but in many cases it is also the cause of motion. sphere or a cylinder rolls over a
surface. irregularities of one surface have little
time to get locked again into the irregularities of the other surface.
 Rolling friction is directly proportional to the normal reaction (R) and inversely proportional to the
radius (r) of the rolling cylinder or wheel.e. A flat block is moving over a horizontal table.
(4) As the portion BC of the curve is parallel to x-axis therefore
kinetic friction does not change with the applied force.
 Rolling friction is often quite small as compared to the sliding friction.
C
B
Fl
O
Fk
Applied force
5.
(iv) Types of kinetic friction
(a) Sliding friction : The opposing force that comes into play when one body is actually sliding over the
surface of the other body is called sliding friction.
inertia of rest has been overcome.
(1) Part OA of the curve represents static friction (Fs ) .3 Graph Between Applied Force and Force of Friction.genius PHYSICS
by Pradeep Kshetrapal
Newton’s Laws of Motion 37
37
(ii) Value of  k depends upon the nature of surface in contact. the force of friction is seen to decrease slightly. Its value increases linearly with the applied force
(3) Beyond A.
The portion BC of the curve therefore represents the kinetic friction
(Fk ) . This represent
limiting friction (Fl ) .
 In rolling the surfaces at contact do not rub each other.
(iii) Kinetic friction is always lesser than limiting friction
Fk  Fl
 k   s
i. This is because once the motion starts actually .
5.
A
Force of friction
(2) At point A the static friction is maximum. That is why heavy loads are
transported by placing them on carts with wheels. it remains
constant. whatever be the applied force. e.
Frolling   r
R
r
 r is called coefficient of rolling friction. Also when motion has actually started. the force of friction comes into play is called rolling friction.
 The velocity of point of contact with respect to the surface remains zero all the times although the
centre of the wheel moves forward.

Sample problems based on fundamentals of friction
Problem 1. the body will remain at rest in the accelerating vehicle until ma   s mg ).38
38 Newton’s Laws of Motion
38
(1) In moving. then what is the maximum force of friction available at the point of contact between the
ladder and the floor
[AIIMS 2002]
(a) 75 N
(b) 50 N
(c) 35 N
(d) 25 N
Maximum force of friction Fl  s R  0.6  1  9. Here the magnitude of pseudo force is less than limiting friction So. If the truck is accelerating at
the rate of 5m/sec2 then frictional force on the block will be
(a) 5 N
(b) 6 N
(c) 5.3. if pedalling is stopped both wheels move by themselves and so
experience force of friction in backward direction.4. when pedalling a bicycle.88 N
When truck accelerates in forward direction at the rate of 5m / s 2 a pseudo force (ma) of 5N works on
block in back ward direction.8  5. The coefficient of static friction is 0. stopped or transferred from one
body to the other. the force of friction is the cause of motion of the
body along with the vehicle (i.6).
Problem 3.
a
smg
ma
From these examples it is clear that without friction motion cannot be started.
Action

(2) In cycling. a person or vehicle pushes the ground
backwards (action) and the rough surface of ground reacts and
exerts a forward force due to friction which causes the motion.]
While
pedalling
Pedalling is
stoped
(3) If a body is placed in a vehicle which is accelerating.
Solution : (a)
Problem 2.
Solution : (a)
If a ladder weighing 250N is placed against a smooth vertical wall having coefficient of friction between it
and floor is 0.
Friction
If there had been no friction there will be slipping and no motion. So. the force exerted by rear wheel on ground makes force of friction act on
it in the forward direction (like walking). Front wheel moving by itself experience force of friction in backward
direction (like rolling of a ball)..88 N
(d) 8 N
Limiting friction  s R  smg  0. If
there had been no friction between body and vehicle the body will not move along with the vehicle. [However.5 N is
applied on the block as shown in the figure.
A block of mass 2 kg is kept on the floor. static
friction works in between the block and the surface of the truck and as we know.3  250  75 N
On the horizontal surface of a truck (  = 0. static friction = Applied
force = 5N. the rear wheel moves by the force communicated to it by pedalling while front wheel moves
by itself.5 N
F
. the frictional force between the block and the floor will be [MP PET 20
(a) 2. If a force F of 2. a block of mass 1 kg is placed.e.

5 Advantages and Disadvantages of Friction.4 × 2 × 9.84 N
As applied force is less than limiting friction. for the given condition static friction will work.
(iii) Brake works on the basis of friction.
Coefficient of static friction  S 
R 20  9 .2 while coefficient of friction
between B and the ground is 0.38
(b) 0. 8
A block of mass M is placed on a rough floor of a lift.
(2) Disadvantages of friction
(i) Friction always opposes the relative motion between any two bodies in contact.
(1) Advantages of friction
(i) Walking is possible due to friction. so a = g]
So force of friction  mg'  0
5.
fBG
Groun
F  f AB  fBG   AB ma g  BG (m A  m B )g
d
= 0.3.
Problem 4.3 (300) × 10
= 200 + 900 = 1100N.5 N. The minimum required force F to start moving B will be
(a) 900 N
(b) 100 N
A
fAB
(c) 1100 N
A
(d) 1200 N
B
F
B
F
Two frictional force will work on block B.2 × 100 × 10 + 0. A horizontal force of 75 N is required to set
the block in motion. When the lift falls freely.
Solution : (d)
A 20 kg block is initially at rest on a rough horizontal surface. The coefficient of friction between the block and the
floor is .8 = 7.
(ii) Friction causes wear and tear of the parts of machinery in contact.5 N and limiting friction = mg = 0. Thus their lifetime reduces. The coefficient of friction between A and B is 0.60
Fl
75

 0 .44
(c) 0. This reduces the efficiency of machine. The coefficient of static friction is
(a) 0. a horizontal force of 60 N is required to keep the block moving
with constant speed.52
(d) 0.
(iv) Writing is not possible without friction. 38 . As shown in figure a horizontal
rope tied to a wall holds it.
Solution : (a)
Problem 6. which causes damage to the machinery. (This is the required minimum force)
Problem 5. After it is in motion. Therefore extra energy
has to be spent in over coming friction.
(ii) Two body sticks together due to friction.
Solution : (c)
A block A with mass 100 kg is resting on another block B of mass 200 kg.
(v) The transfer of motion from one part of a machine to other part through belts is possible by friction.
. So. the block is pulled horizontally on the floor.
Static friction on a body = Applied force = 2.84 N
(d) 10 N
Applied force = 2.genius PHYSICS
by Pradeep Kshetrapal
Newton’s Laws of Motion 39
39
Solution : (a)
(b) 5 N
(c) 7. What will be the force of
friction
(a)  Mg
(b)  Mg/2
(c) 2 Mg
(d) None of these
When the lift moves down ward with acceleration 'a' then effective acceleration due to gravity
g' = g – a
g'  g  g  0 [As the lift falls freely.
(iii) Frictional force result in the production of heat.

(5) By using ball bearing.
As well as    i.
By definition angle  is called the angle of friction
R
S
F

tan  
F
P
R
F

tan  = 
[As we know
 ]
R
mg
or
  tan 1 ( )
Hence coefficient of limiting friction is equal to tangent of the angle of friction.
R
F
In limiting condition F  mg sin 
and
R  mg cos 
mg sin 
F
So
 tan 
R

F
F

   tan   tan  [As we know
   tan  ]
R
R
Thus the coefficient of limiting friction is equal to the tangent of angle of repose.
Sample problems based on angle of friction and angle of repose

mg
mg cos 
..e.
5.9 Angle of Repose.
Also we can increase friction by throwing some sand on slippery ground. S = mg
Hence the range of S can be given by.6 Methods of Changing Friction. angle of repose = angle of friction.
In the above figure resultant force S  F 2  R 2
S  (mg )2  (mg )2
S  mg  2  1
when there is no friction (  0) S will be minimum i. In the manufacturing of tyres.
(2) By lubrication.40
40 Newton’s Laws of Motion
40
5.
Angle of repose is defined as the angle of the inclined plane with horizontal such that a body placed on it is
just begins to slide.7 Angle of Friction.e.
(3) By proper selection of material.
Angle of friction may be defined as the angle which the resultant of limiting friction and normal reaction
makes with the normal reaction.8 Resultant Force Exerted by Surface on Block.
By definition  is called the angle of repose.
5.
mg  S  mg  2  1
5.
(4) By streamlining the shape of the body.
synthetic rubber is preferred because its coefficient of friction with the road is larger.
We can reduce friction
(1) By polishing.

then following four
situations are possible
(i) When there is no friction
(a) The body A will move on body B with acceleration (F/m).45J
(d) 1.
2
Problem 18.
aA  F / m
A
L
M
m
M
m
F
B
F
A
B
L
(b) The body B will remain at rest
aB  0
(c) If L is the length of B as shown in figure A will fall from B after time t
1 2


 As s  2 a t and a  F/m 


(ii) If friction is present between A and B only and applied force is less than limiting friction (F < Fl)
(F = Applied force on the upper body./Med. A block of mass 1 kg slides down on a rough inclined plane of inclination 60 o starting from its top.82 J
(b) 4.2  50  9.
When a body A of mass m is resting on a body B of mass M then two
conditions are possible
(1) A force F is applied to the upper body.e.8 m/s2)
[AFMC 2000.
[CBSE PMT 1999.
Free body diagram of B
MaB
FK
B
. Fk = Kinetic friction between A and
t
2L

a
2mL
F
B)
(a) The body A will not slide on body B till F  Fl i.8  1  98 J .e.e. BHU 2001]
(c) 56 J
(d) 34 J
5. Here force of kinetic friction  k mg will oppose the motion of A while will cause the motion of B. If
Solution : (c)
the coefficient of kinetic friction is 0. Fl = limiting friction between A and B. 8  = 2.94 J
(c) 2.e. If the coefficient of friction between their
Solution : (a)
surfaces is 0.2.13 Motion of Two Bodies One Resting on the Other. then work done against friction
is (Take g = 9.5 and length of the plane is 1 m.96 J
1
W  mg cos  .S  0 . of applied force) but with different
acceleration. 2000. then work done against friction is
(a) 98 J
(b) 72J
W  mgS  0. a A 
aA
Note
:
Fk  M a B
maA
i. KCET (Engg.46
46 Newton’s Laws of Motion
46
Problem 17. A block of mass 50 kg slides over a horizontal distance of 1 m.
(2) A force F is applied to
the lower body
We will discuss above two cases one by one in the following manner :
(1) A force F is applied to the upper body. AIIMS 2000. F   s mg
F
M m
(iii) If friction is present between A and B only and applied force is greater than limiting friction (F
(b) Combined system (m + M) will move together with common acceleration a A  a B 
> Fl)
In this condition the two bodies will move in the same direction (i.
A
Fk
F

aB 
aB 
Fk
M
 k mg
M
As both the bodies are moving in the same direction.
F  Fk  m a A
F  Fk
m
(F   k mg )

m
Free body diagram of A
i.) 2001]
(a) 9.45 J. 5  1  9 .

a A   k g
A
maA
Fk
Note
:
[ F   k mg ]
i.
MaB
FK
B
Fl
(2) A force F is applied to the lower body. Here force of kinetic friction  k mg will oppose the
motion of B while will cause the motion of A. Fk = kinetic friction between A and B)
B will move only if Fk  Fl and then Fk  Fl  M a B
However if B does not move then static friction will work (not limiting
friction) between body B and the floor i.genius PHYSICS
by Pradeep Kshetrapal
Newton’s Laws of Motion 47
47
Acceleration of body A relative to B will be a  a A  a B 
So. A will fall from B after time t 
2L

a
MF   k mg (m  M )
mM
2 m ML
MF   k mg (m  M )
(iv) If there is friction between B and floor
(where Fl    (M  m ) g = limiting friction between B and floor. A will move backwards with acceleration (F/M) and so will fall from it in time t.
ma A  k mg
F  Fk  Ma B
Free body diagram of A
i.
 t
2L

a
2 ML
F
(ii) If friction is present between A and B only and F < Fl
(where F = Pseudo force on body A and Fl = limiting friction between body A and B)
F
(a) Both the body will move together with common acceleration a 
M m
mF
(b) Pseudo force on the body A.e. friction force = applied force (= Fk) not
Fl .
F
a B    and a A  0
M
(b) As relative to B. F   ma 
and Fl   s mg
mM
mF
(c) F   Fl 
  s mg  F   s (m  M ) g
mM
F
So both bodies will move together with acceleration a A  a B 
if F   s [m  M ] g
mM
(iii) If friction is present between A and B only and F > Fl
(where Fl = s (m + M)g = limiting friction between body B and surface)
Both the body will move with different acceleration.e. then following four
situations are possible
(i) When there is no friction
(a) B will move with acceleration (F/M) while A will remain at rest (relative to ground) as there is no
pulling force on A. a B 
M
As both the bodies are moving in the same direction
Acceleration of body A relative to B will be
Free body diagram of B
MaB
FK
B
F
.e.

3 N
(d) It will not slide for any F
A
B will begin to slide on A if Pseudo force is more than limiting friction
 F 
 F 
F'  Fl  m 
  s R  m 
  0 . A body A of mass 1 kg rests on a smooth surface.764 N
m

M


m  M 
Problem 21. A just slips on B when a
force of 12 N is applied on A.
If F is the maximum value of force applied on lower body such that both body move together
It means Pseudo force on upper body is just equal to limiting friction
 F   4 
F'  Fl  m 

 F  12  F  36 N .
Sample problems based on body resting on another
Problem 19. B will begin to slide on A if A is pulled
with a force greater than
Solution : (a)
(a) 1.764 N
(b) 0. A 4 kg block A is placed on the top of a 8 kg block B which rests on a smooth table. limiting friction between A and B. A will move backwards and will fall it after time
t
2L

a
2 ML
F   k g(m  M )
(iv) If there is friction between B and floor : The system will move only if F  Fl' then replacing F by
F  Fl . A block A of mass 2 kg rests on another block B of mass 8 kg which rests on a horizontal floor.
m  M  4 8 
Problem 20.5(2  8)10  50 N
but the applied force is 25 N so the lower block will not move i.5. the force of friction between A and B is
[IIT-JEE 1993]
(a) Zero
Solution : (a)
(b) 3.e.
A
B
2kg
8kg
Surface
25 N
.2.R  0. while that between B and floor is 0.
is
(a) 12 N
Solution : (c)
(b) 24 N
(c) 36 N
(d) 48 N
Maximum friction i. The entire case (iii) will be valid.15. The
coefficient of friction between A and B is 0.1764 N
B
(c) 0.0 N
(d) 49 N
Limiting friction between the block B and the surface
FBS  BS . The coefficient of static friction between A and B is 0.2 kg is placed over A as
shown. Then the maximum horizontal force on B to make both A and B move together.e. Hence there will be no
force of friction between A and B. Fl = 12 N. Another body B of mass 0.
there is no pseudo force on upper block A.14 Motion of an Insect in the Rough Bowl.
5.15 mg F  1.
However if F  Fl the system will not move and friction between B and floor will be F while between A
and B is zero.9 N
(c) 5.48
48 Newton’s Laws of Motion
48
 F   k g(m  M ) 
a  a A  a B  

M


Negative sign implies that relative to B. When a horizontal
force of 25 N is applied on the block B.5 m  M  g  0.

If the coefficient of friction is 0. Its velocity on reaching the bottom is v.49
(d) 0.
(b) 1 : 2
11.6 sin 45o
7. The coefficient of friction will be
(a) 0.
(c)
FA  FB  FC  FD
(d)
FA  FB  FC  FD
(b) 0. one should take smaller steps because of the
[BHU 1999]
(a) Friction of ice is large
(b) Larger normal reaction
(c) Friction of ice is small
(d) Smaller normal reaction
Two bodies having the same mass.25
(c) 0.75
(d) 0. The coefficient of friction between the body and the inclined plane is
[CBSE PMT 1990]
(a) 0. If the
1
coefficient of friction is . because friction
(a) Resists motion
(b) Causes wear and tear
(c) Depends upon the nature of materials
(d) Operating in this case is sliding friction
The angle between frictional force and the instantaneous velocity of the body moving over a rough surface is
(a) Zero
(b) /2
(c) 
(d) Equal to the angle of friction
What happens to the coefficient of friction.
8.2.
C
B
g
2
(b)
g
2 2
(c)
g  1
1  
2  2
(d)
g  1
1  
2  2
A block moves down a smooth inclined plane of inclination .
The coefficients of static and sliding friction are 0.
(c) 2 : 1
(b) 0.8 sin 45o
(d) 9.) 1999]
(a) 1 : 1
9. a body slides down a 45 o inclined plane in twice the time it takes to slide down the same distance in the
absence of friction.56
56 Newton’s Laws of Motion
56
(a) 19. If it slides down a rough
inclined plane of same inclination its velocity on reaching the bottom is v/n.5 and 0.
(d) 1 : 4
Starting from rest. when the normal reaction is halved
(a) Halved
(b) Doubled
(c) No change
(d) Depends on the nature of the surface
What can be inferred regarding the limiting frictional force in the following four figures
R
R
R
R
A
mg
mg
(a) FA  FB  FC  FD
14. The acceleration of the body is
(a) 6 m / sec 2
16.8 cos 45o
To avoid slipping while walking on ice.
12.1
(b) 4.92 m / sec 2
(d) 1 m / sec 2
A particle is projected along a line of greatest slope up a rough plane inclined at an angle of 45 o with the horizontal.80
Brakes of very small contact area are not used although friction is independent of area. The coefficient
of friction  is given by
1

(a)   tan  1  2 
 n 
1

(b)   cot  1  2 
 n 
1
1 2

(c)   tan  1  2 
 n 
1
(d)
1 2

  cot  1  2 
 n 
. then the retardation is
2
(a)
17.6 cos 45o
(c) 9. the forces of friction that come into play when they are in motion will be in the ratio
[EAMCET (Med.4 respectively.
(b)
D
A force of 98 Newton is required to drag a body of mass 100 kg on ice.
FA  FB  FC  FD
mg
mg
A 60 kg body is pushed with just enough force to start it moving across a floor and the same force continues to act afterwards.98
15.
(b) 19.89
(c) 0. where n is a number greater than 0.9 m / sec 2
(c) 3. 2 kg each have different surface areas 50 m 2 and 100 m 2 in contact with a horizontal plane.
18.33
10.
13.

The coefficient of static friction
between the block and the plane is 0. 8  3 N
(d) 0. The frictional force on the block is
[IIT-JEE 1980]
(a) 9. then the minimum force required to
move the body along the surface will be
(a) W tan
26. The direction of friction on B due to A is
(a) Zero
(b) To the left
BA
F
(c) Upwards
(d) Downwards
21. The force acting on
the body down the plane in this position is
(a)
Mg
(b)
Mg
3
2
Mg
3
(c)
(d)
Mg
10
 Advance level
22.5.
(b) 30o
(b) W cos
(c) W sin
(d) W cos
A block of mass M is placed on a rough horizontal surface as shown in the figure.
(b) 40 m
(c) 72 m
(d) 20 m
All the surfaces shown in the figure are rough.8 N
A body of weight W is lying at rest on a rough horizontal surface.
A body of mass M just starts sliding down an inclined plane (rough) with inclination . If the coefficient of static friction between the
tyres and the road is 0.8 N
25.
A plane is inclined at an angle  with the horizontal. If the coefficient of friction is .) 2000]
(a) 15o
24. The coefficient of friction is . A force F = Mg acts on the block. The block can be pushed along the surface only when
(a) tan   
F = Mg
(b) cot   
(c)

M
tan  / 2  
(d) cot  / 2  
27.8  3 N
(c) 9 .7  9 .
Consider a car moving along a straight horizontal road with a speed of 72 km/hr. If the normal reaction is twice that of the resultant
downward force along the incline. the angle between the inclined plane and the horizontal is
[EAMCET (Engg.
(d) 60o
(b) 0.
Consider the following statements
Assertion (A) : It is difficult to move a cycle along the road with its brakes on.
A body is sliding down an inclined plane having coefficient of friction 0. the shortest distance in which the car can be stopped is (g  10 m / s 2 )
(a) 30 m
20. such that tan = 1/3.5.
Reason (R) : Sliding friction is greater than rolling friction. then the
minimum force that has to be applied parallel to the inclined plane to make the body just move up the inclined plane is
(a) mg sin 
(b)  mg cos
(c)  mg cos – mg sin
(d)  mg cos + mg sin
.
Of these statements
[AIIMS 2002]
(a) Both A and R are true and the R is a correct explanation of the A
(b) Both A and R are true but the R is not a correct explanation of the A
(c) A is true but the R is false
(d) Both A and R are false
(e) A is false but the R is true
23. It is inclined
to the vertical at an angle .7. A body of mass m rests on it. If the angle of friction is .genius PHYSICS
by Pradeep Kshetrapal
Newton’s Laws of Motion 57
57
19.7  9.
(c) 45o
A block of mass 2 kg rests on a rough inclined plane making an angle of 30 o with the horizontal.

When the force acting on X is 12N. If M1
= 10 kg.
Two masses 10 kg and 5 kg are connected by a string passing over a pulley as shown.3 kg
(c) 32.15 and F = 40 N. The coefficient of friction
between the two blocks is  1 and that between the block of mass M and horizontal surface is  2 .5 kg
(d) 44. What maximum horizontal
force can be applied to the lower block so that the two blocks move without separation
(a) (M + m) (2  1 )g
m
(b) (M – m) (2  1 )g
M
(c) (M – m) (2  1 )g
(d) (M + m) (2  1 )g
29.
Two blocks of mass M1 and M2 are connected with a string which passes over a smooth pulley. As shown in the figure.000 kg pulls a coach of mass 40.000 kg. The mass
M1 is placed on a rough inclined plane as shown in the figure.3 kg
33. the coefficient of friction between the road and the tyres is .7 kg
(b) 23.
A block of mass m is placed on another block of mass M which itself is lying on a horizontal surface. A force F acts on the block M1. If the coefficient of friction be 0.
(b)

(c)
1

(d)
1

An engine of mass 50.58
58 Newton’s Laws of Motion
58
28. 2 = 0.
(b) 2 ms 2
(c) 1 ms 2
(d) None of these
A block X of mass 4 kg is lying on another block Y of mass 8 kg. The coefficient of friction between the block
and the inclined plane is . 1 = 0.500 N.
A block of mass M1 is placed on a slab of mass M2.5.6
(c) 0. Take g = 10 ms–2.6
32. The slab lies on a frictionless horizontal surface.
In the above problem if F = 100 N. then the acceleration of the engine is
.
A car starts from rest to cover a distance s. If there is a resistance of 1 N per 100 kg acting on both the engine
and the coach. The coefficient of static
friction between the block and slab is 1 and that of dynamic friction is 2. what will be the acceleration with which the slab will move
(a) 5 ms 2
31. The minimum time in
which the car can cover the distance is proportional to
(a) 
35. and if the driving force of the engine be 4. M2 = 30 kg. What should be the maximum mass M2 so that block M1 slides downwards
(a) M 2  M1(sin    cos  )
(b) M 2  M1 (sin    cos  )
(c) M 2  M1 /(sin    cos  )
(d) M 2  M1 /(sin    cos  )
34. The force F in Newton necessary to make both X and Y move simultaneously will be
(a) 36
(b) 3.15. block
X is on the verge of slipping on Y. what will be the acceleration with which the slab will move
(a) 5 ms 2
(b) 2 ms 2
(c) 1 ms 2
(d) Zero
30. then
the minimum weight that may be placed on 10 kg to stop motion is
(a) 18.36
(d) 3.

where k  1.
(d) 4 m/s2
(b) tan  = 2 tan 
(c) tan  = 3 tan 
(d) tan  = 3 tan 
A body is on a rough horizontal plane.
(b) 2 m/s2
(b)   
(c)   
(d)  can take up any value
In the arrangement shown W1  200 N . A force is applied to the body direct towards the plane at an angle  with the vertical.The mass of the plane is 10 4 kg and the
coefficient of friction between the plane and the ground is 0. The coefficient of friction between the surface of the block
is 1/4.
Mg
 1
 (M  m )g
 1
A body slides over an inclined plane forming an angle of 45° with the horizontal. The plane accelerates uniformly during take off. The block W1 just slides under the
block W2
(a) A pull of 50 N is to be applied on W1
(b) A pull of 90 N is to be applied on W1
(c) Tension in the string AB is 10 2 N
(d) Tension in the string AB is 20 2 N
41. The run of the ground is 100m.genius PHYSICS
by Pradeep Kshetrapal
Newton’s Laws of Motion 59
59
(a) 0. A light string passing over the smooth light pulley is used to connect A and B as shown.08 m / s 2
36.
A board of mass m is placed on the floor and a man of mass M is standing on the board as shown. The coefficient of friction between the body and the plane has a value
(a)   0.25 for all surfaces in contact. The coefficient of friction
between the board and the floor is .500 N
An aeroplane requires for take off a speed of 72 km/h.   0. then tension in the coupling is
(a) 2. If A is dragged with a force F then for both A and B to move with a uniform speed we
have
(a) F  (M  m)g
(b) F  mg
(c)
F  (3 M  m)g
(d) F  (3m  M)g
44.25
(d)   0.5
43.04 m / s 2
(d) None of these
(c) 500 N
(d) 1000 N
In the above question. W2  100 N . If 
is angle of friction and  is the angle which incline makes with the horizontal then
(a) tan  = tan 
39. If  is
the angle of friction then for the body to move along the plane
(a)   
40.
(c) 0.
(c) 3 m/s2
The force required to just move a body up an inclined plane is double the force required to just prevent it from sliding down.732.
(b) Zero
(b) 1.000 N
37. The distance x travelled by the body in time t is
described by the equation x  kt 2 .
A force of 100 N is applied on a block of mass 3 kg as shown in figure. What is the
acceleration of the plane
(a) 1 m/s2
38. The maximum force that the can exert on he rope so that the board does not slip on the
floor is
(a) F  (M  m)g
(b) F  mg
(c)
F
(d) F 
42.75
Two blocks A and B of masses m and M respectively are placed on each other and their combination rests on a fixed horizontal
surface C. The coefficient of sliding
friction between all surfaces in contact is .
(b)   1
(c)
  0.2. The friction force acting on the block is
(a) 15 N downwards
(b) 25 N upwards
(c)20 N downwards (d)20 N upwards
F = 100 N
.